Question

This may be a trivial question. Usually, I need to define local std::vectors like vector<int>, vector<double> etc to store some intermediate values for a vector<A> where A is a self-defined class. These local vectors are supposed to have the same size as the vector<A>.

The problem is, when I write some for to assign values from vector<A> to the local vectors, is it bad practice to write it like (PS: the size of vect_a remains unchanged in the loop.)

vector<A> vect_a;
vector<int> local_vect;
...
for (vector<A>::size_type ix = 0; ix != vect_a.size(); ++ix) {
    local_vect[ix] = vect_a[ix].get_mem_var();
    ...
}

where get_mem_var() simply returns the int mem_var of A. Specifically, my concern is, local_vect is of vector<int>, not vector<A>, but it still uses ix, which is of vector<A>::size_type instead of vector<int>::size_type.

Sometimes I make it a bit more complex, like

for (struct {vector<A>::size_type ix; vector<int>::size_type iy;}
        gi = {0, 0}; gi.ix != vect_a.size(); ++gi.ix, ++gi.iy) {
    local_vect[gi.iy] = vect_a[gi.ix].get_mem_var();
    ...
}

But it feels redundant and unnecessary. I've heard size_type is often simply size_t, and using vector<A>::size_type ix for indexing vector<int> local_vect may just be OK. But is this bad practice in some way? A more general question is, is it OK to use vector<A>::size_type ix to index vector<A> and vector<B> (A and B can be built-in or user-defined) if we know vector<A> and vector<B> are supposed to have the same size?

Answer 1

实际上，std::vector<T>::size_type可能与任何T相同，因此您无需担心这一点。

您可以使用std::transform来避免担心这一点，例如：

vector<A> vect_a;
vector<int> local_vect;

std::transform(std::begin(vect_a), std::begin(vect_b), std::begin(local_vect),
               [] (const A& a) { return a.get_mem_var(); });

使用`vector <a>::size_type ix` to index `vector<b>`

1 个答案: